Week 2: Linear models and causal inference

Workspace setup:

library(tidyverse)
library(cowplot)
library(brms)
library(tidybayes)
library(patchwork)
library(here)

As we develop more useful models, we’ll begin to practice the art of generating models with multiple estimands. An estimand is a quantity we want to estimate from the data. Our models may not themselves produce the answer to our central question, so we need to know how to calculate these values from the posterior distributions.

This is going to be different from prior regression courses (PSY 612), where our models were often designed to give us precisely what we wanted. For example, consider the regression:

\[ \hat{Y} = b_0 + b_1(D) \] Where \(Y\) is a continuous outcome and \(D\) is a dummy coded variable (0 = control; 1 = treatment).

• What does \(b_0\) represent?
• What does \(b_1\) represent?
• How would you calculate or estimate the means of both groups from this model?

Categories

If you were to fit the dummy variable model in R, your formula would be:

y ~ 1 + D

or

y ~ D # intercept is implied

Forget dummy codes. From here on out, we will incorporate categorical causes into our models by using index variables. An index variable contains integers that correspond to different categories. The numbers have no inherent meaning – rather, they stand as placeholders or shorthand for categories.

To fit these models in R, we’re going to force the model to drop the intercept.

y ~ 0 + I

Mathematical model

Let’s write a mathematical model to express weight in terms of sex.

\[\begin{align*} w_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{SEX[i]} \\ \alpha_j &\sim \text{Normal}(130, 20)\text{ for }j = 1..2 \\ \sigma &\sim \text{Uniform}(0, 25) \end{align*}\]

Fitting the model using brms()

m1 <- brm(
  data = d,
  family = gaussian,
1  bf(weight ~ 0 + a,
2     a ~ 0 + sex,
3     nl = TRUE),
4  prior = c(prior( normal(130, 20), class=b, nlpar = a),
            prior( uniform(0, 25),  class=sigma, ub=25)),
  iter = 2000, warmup = 1000, seed = 3, chains=1,
  file = here("files/models/22.1")
)

posterior_summary(m1)

            Estimate  Est.Error        Q2.5       Q97.5
b_a_sex1    92.27947 0.91187785    90.50115    94.05188
b_a_sex2   107.22803 0.91554891   105.57421   109.04851
sigma       12.19829 0.46172983    11.34413    13.13513
lprior     -13.47705 0.09932469   -13.66427   -13.27516
lp__     -1390.30623 1.23903576 -1393.80224 -1388.90647

Here, we are given the estimates of the parameters specified in our model: the average weight of women (b_a_sex1) and the average weight of men (b_a_sex2). But our question is whether these average weights are different. How do we get that?

Calculate the contrast

We can create two plots. One is the posterior distributions of average female and male weights and one is the average difference.

p1 <- post %>% 
  pivot_longer(starts_with("b")) %>% 
  mutate(sex = ifelse(str_detect(name, "1"), "female", "male")) %>% 
  ggplot(aes(x=value, color = sex)) +
  geom_density(linewidth = 2) +
  labs(x = "weight(lbs)") 

p2 <- post %>% 
  mutate(diff_fm = b_a_sex1 - b_a_sex2) %>%
  ggplot(aes(x=diff_fm)) +
  geom_density(linewidth = 2) +
  labs(x = "difference in weight(lbs)") 

( p1 | p2)

Expected values vs predicted values

A note that the distributions of the mean weights is not the same as the distribution of weights period. For that, we need the posterior predictive distributions. Here are two methods for getting predicted values.

Method 1: simulate using the rnorm()function. This is more intuitive and will help you mentally reconnect the parameters of your statistical model with the causal model you started with. But there’s more room for human error.

pred_f  <- rnorm(nrow(post), mean = post$b_a_sex1, sd = post$sigma )
pred_m  <- rnorm(nrow(post), mean = post$b_a_sex2, sd = post$sigma )

pred_post = data.frame(pred_f, pred_m) %>%
  mutate(diff = pred_f-pred_m)
head(pred_post)

    pred_f    pred_m        diff
1 82.39016 118.45218 -36.0620188
2 90.93993  88.69816   2.2417732
3 88.84118 129.20769 -40.3665072
4 79.43751 118.08744 -38.6499322
5 98.51033  97.99357   0.5167583
6 80.59447 114.24528 -33.6508099

Expected values vs predicted values

Method 2: use the predicted_draws() function. This is less intuitive, but there’s less room for you to make a mistake.

nd = distinct(d, sex)
pred_all = predicted_draws(object=m1, newdata=nd) %>% 
  ungroup %>% select(-.row) %>% 
  pivot_wider(names_from = sex, names_prefix = "sex", values_from = .prediction) %>% 
  mutate(diff = sex1-sex2)
head(pred_all)

# A tibble: 6 × 6
  .chain .iteration .draw  sex2  sex1   diff
   <int>      <int> <int> <dbl> <dbl>  <dbl>
1     NA         NA     1  112. 113.    1.09
2     NA         NA     2  109.  83.3 -25.7 
3     NA         NA     3  118. 101.  -17.1 
4     NA         NA     4  105.  90.4 -14.7 
5     NA         NA     5  122.  70.8 -51.1 
6     NA         NA     6  104.  76.0 -28.0 

exercise

In the rethinking package, the dataset milk contains information about the composition of milk across primate species, as well as some other facts about those species. The taxonomic membership of each species is included in the variable clade; there are four categories.

1. Create variable in the dataset to assign an index value to each of the 4 categories.
2. Standardize the milk energy variable (kcal.per.g). ¹
3. Write a mathematical model to express the average milk energy (in standardized kilocalories) in each clade.

solution

data("milk", package="rethinking")
str(milk)

'data.frame':   29 obs. of  8 variables:
 $ clade         : Factor w/ 4 levels "Ape","New World Monkey",..: 4 4 4 4 4 2 2 2 2 2 ...
 $ species       : Factor w/ 29 levels "A palliata","Alouatta seniculus",..: 11 8 9 10 16 2 1 6 28 27 ...
 $ kcal.per.g    : num  0.49 0.51 0.46 0.48 0.6 0.47 0.56 0.89 0.91 0.92 ...
 $ perc.fat      : num  16.6 19.3 14.1 14.9 27.3 ...
 $ perc.protein  : num  15.4 16.9 16.9 13.2 19.5 ...
 $ perc.lactose  : num  68 63.8 69 71.9 53.2 ...
 $ mass          : num  1.95 2.09 2.51 1.62 2.19 5.25 5.37 2.51 0.71 0.68 ...
 $ neocortex.perc: num  55.2 NA NA NA NA ...

milk$clade_id <- as.integer(milk$clade)
milk$clade_id <- as.factor(milk$clade_id)
milk$K <- rethinking::standardize(milk$kcal.per.g)

\[\begin{align*} K_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{\text{CLADE}[i]} \\ \alpha_i &\sim \text{Normal}(0, 0.5) \text{ for }j=1..4 \\ \sigma &\sim \text{Exponential}(1) \\ \end{align*}\]

Exercise: Now fit your model using brms(). It’s ok if your mathematical model is a bit different from mine.

solution

m2 <- brm(
  data=milk,
  family=gaussian,
  bf(K ~ 0 + a,
     a ~ 0 + clade_id,
     nl = TRUE),
  prior = c( prior(normal(0,.5), class=b, nlpar=a),
             prior(exponential(1), class=sigma)),
  iter = 2000, warmup = 1000, seed = 3, chains=1,
  file = here("files/models/22.2")
)

posterior_summary(m2)

                 Estimate Est.Error         Q2.5        Q97.5
b_a_clade_id1  -0.4641202 0.2256089  -0.89176596  -0.03325312
b_a_clade_id2   0.3417955 0.2276394  -0.12749964   0.78516752
b_a_clade_id3   0.6493047 0.2857904   0.09024468   1.18518255
b_a_clade_id4  -0.5387150 0.2962131  -1.10847821   0.03517569
sigma           0.7991983 0.1178092   0.61678499   1.07557053
lprior         -4.3341786 1.1496078  -6.91138553  -2.49326609
lp__          -38.8228118 1.5985564 -42.59133454 -36.64554417

exercise

Plot the following distributions:

• Posterior distribution of average milk energy by clade.
• Posterior distribution of predicted milk energy values by clade.

solution

post <- as_draws_df( m2 )
post %>% 
  pivot_longer(starts_with("b")) %>% 
  mutate(
    name = str_extract(name, "[0-9]"),
    name = factor(name, labels = levels(milk$clade))) %>% 
  ggplot(aes(x = value, color = name)) +
  geom_density(linewidth = 2) +
  labs(title = "Posterior distribution of expected milk energy") +
  theme(legend.position = "bottom")

solution

nd = distinct(milk, clade_id)
ppd <- posterior_predict( m2, newdata = nd)
ppd = as.data.frame(ppd)
names(ppd) = levels(milk$clade)
ppd %>% 
  pivot_longer(everything()) %>% 
  ggplot(aes(x = value, color = name)) +
  geom_density(linewidth = 2) +
  labs(title = "Posterior predictive distribution of predicted milk energy") +
  theme(legend.position = "bottom")

Plotting with brms and tidybayes

as_draws_df(m2) %>% 
  pivot_longer(starts_with("b")) %>% 
  mutate(
    clade = str_extract(name, "[0-9]"),
    clade = as.numeric(clade),
    clade = factor(clade, labels=levels(milk$clade))
  ) %>%
  ggplot(aes(y = clade, x = value)) +
  stat_halfeye() +
  labs(x="mean", y=NULL)

Combining index variables and slopes

Let’s return to the weight example. What if we want to control for height?

\[\begin{align*} w_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{S[i]} + \beta_{S[i]}(H_i-\bar{H})\\ \alpha_j &\sim \text{Normal}(130, 20)\text{ for }j = 1..2 \\ \beta_j &\sim \text{Normal}(0, 25)\text{ for }j = 1..2 \\ \sigma &\sim \text{Uniform}(0, 50) \end{align*}\]

d$height_c <- d$height - mean(d$height)
m3 <- brm(
  data = d,
  family = gaussian,
  bf(weight ~ 0 + a + b*height_c,
     a ~ 0 + sex,
     b ~ 0 + sex,
     nl = TRUE),
  prior = c(prior(normal(130, 20), class = b, nlpar = a),
            prior(normal(  0, 25), class = b, nlpar = b),
            prior(uniform( 0, 50), class = sigma, lb=0, ub=50)),
  iter = 2000, warmup = 1000, seed = 3, chains=1,
  file = here("files/models/22.3")
)

Plot the slopes from the posterior

post = as_draws_df(m3) %>% 
  pivot_longer(starts_with("b"),
               names_to = c("parameter","sex"),
               names_sep = 3) %>% 
  mutate(sex = str_extract(sex, "[0-9]")) %>% 
  pivot_wider(names_from = parameter, values_from = value)
xlabs = seq(4,6,by=.5)

d %>% 
  ggplot(aes(x=height_c, y=weight)) +
  geom_point(aes(color=sex)) + 
  geom_abline(
    aes(intercept=b_a, slope=b_b, color=sex), 
    data=post[1:40, ],
    alpha=.2
  ) + 
  scale_x_continuous("height(feet)",
                     breaks=xlabs-mean(d$height),
                     labels=xlabs) +
  facet_wrap(~sex)

exercise

Return to the milk data. Write a mathematical model expressing the energy of milk as a function of the species body mass (mass) and clade category. Be sure to include priors. Fit your model using brms().

solution

\[\begin{align*} K_i &\sim \text{Normal}(\mu_i, \sigma) \\ \mu_i &= \alpha_{\text{CLADE}[i]} + \beta_{\text{CLADE}[i]}(M-\bar{M})\\ \alpha_i &\sim \text{Normal}(0, 0.5) \text{ for }j=1..4 \\ \beta_i &\sim \text{Normal}(0, 0.5) \text{ for }j=1..4 \\ \sigma &\sim \text{Exponential}(1) \\ \end{align*}\]

milk$mass_c = milk$mass- mean(milk$mass)

m4 <- brm(
  data = milk,
  family = gaussian,
  bf(K ~ 0 + a + b*mass_c,
     a ~ 0 + clade_id,
     b ~ 0 + clade_id,
     nl = TRUE),
  prior = c(prior( normal(0,.5), class=b, nlpar = a),
            prior( normal(0,.5), class=b, nlpar = b),
            prior( exponential(1),  class=sigma)),
  iter = 2000, warmup = 1000, seed = 3, chains=1,
  file = here("files/models/22.4")
)